Assume that a group is a group extension given as a
short exact sequence of groups : with
kernel, , and quotient, . If the kernel, , is a complete group then the extension splits: is isomorphic to the
direct product, . A proof using
homomorphisms and exact sequences can be given in a natural way: The action of (by
conjugation) on the
normal subgroup, , gives rise to a group homomorphism, . Since and has trivial center the homomorphism is surjective and has an obvious section given by the inclusion of in . The kernel of is the
centralizer of in , and so is at least a
semidirect product, , but the action of on is trivial, and so the product is direct. This can be restated in terms of elements and internal conditions: If is a normal, complete
subgroup of a group , then is a direct product. The proof follows directly from the definition: is centerless giving is trivial. If is an element of then it induces an automorphism of by conjugation, but and this conjugation must be equal to conjugation by some element of . Then conjugation by is the identity on and so is in and every element, , of is a product in . == References ==